Question

One of the square faces of a metal slab of side $50\text{cm}$ and thickness $20\text{cm}$ is rigidly fixed on a horizontal surface. If a tangential force of $30\text{N}$ is applied to the top face, then the displacement (in $\text{m}$) of the top face is :
[The shear modulus of the material is $4\times {10}^{10}{\text{N/m}}^2$]

• A. $4\times {10}^{-12}$
• B. $4\times {10}^{-10}$
• C. $6\times {10}^{-10}$
• D. $8\times {10}^{-10}$

Answer 1

The correct option is C $6\times {10}^{-10}$


Shear strain $(\in )=\tan \theta =\frac{x}{l}$
where $x$ is the displacement of face and $l$ is the thickness.
Shear stress $\sigma =\frac{F}{A}$
We know, Shear modulus $\left(\eta \right)=\frac{\sigma }{\in }$
Hence, $x=\frac{F.l}{A.\eta }$

Given, $F=30\text{N}$
$A=50\times 50\times {10}^{-4}{\text{m}}^2$
$\eta =4\times {10}^{10}{\text{N/m}}^2$
$\Rightarrow x=\frac{30\times 20\times {10}^{-2}}{50\times 50\times {10}^{-4}\times 4\times {10}^{10}}$
$\Rightarrow x=\frac{6}{25\times 4\times {10}^{-2}\times {10}^{10}}$
i.e $x=6\times {10}^{-10}\text{m}$